Simplify the following expression: $\dfrac{132k^5}{110k^5}$ You can assume $k \neq 0$.
Answer: $ \dfrac{132k^5}{110k^5} = \dfrac{132}{110} \cdot \dfrac{k^5}{k^5} $ To simplify $\frac{132}{110}$ , find the greatest common factor (GCD) of $132$ and $110$ $132 = 2 \cdot 2 \cdot 3 \cdot 11$ $110 = 2 \cdot 5 \cdot 11$ $ \mbox{GCD}(132, 110) = 2 \cdot 11 = 22 $ $ \dfrac{132}{110} \cdot \dfrac{k^5}{k^5} = \dfrac{22 \cdot 6}{22 \cdot 5} \cdot \dfrac{k^5}{k^5} $ $\phantom{ \dfrac{132}{110} \cdot \dfrac{5}{5}} = \dfrac{6}{5} \cdot \dfrac{k^5}{k^5} $ $ \dfrac{k^5}{k^5} = \dfrac{k \cdot k \cdot k \cdot k \cdot k}{k \cdot k \cdot k \cdot k \cdot k} = 1 $ $ \dfrac{6}{5} \cdot 1 = \dfrac{6}{5} $